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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9982 | Accepted: 5724 |
Description
The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices. 1033The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.
1733
3733
3739
3779
8779
8179
Input
Output
Sample Input
3 1033 8179 1373 8017 1033 1033
Sample Output
6 7 0
#include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define maxn 100001 const int inf=0x7fffffff; //无限大 const int MAXN = 10000; bool flag[MAXN]; int primes[MAXN], pi; struct point { int x; int y; }; void GetPrime_1() { int i, j; pi = 0; memset(flag, false, sizeof(flag)); for (i = 2; i < MAXN; i++) if (!flag[i]) { primes[i] = 1;//素数标识为1 for (j = i; j < MAXN; j += i) flag[j] = true; } } int vis[maxn]; int main() { GetPrime_1(); int t; cin>>t; while(t--) { memset(vis,0,sizeof(vis)); int n,m; cin>>n>>m; vis[n]=1; queue<point> q; q.push((point){n,0}); int flag1=0; while(!q.empty()) { point now=q.front(); if(now.x==m) { flag1=now.y; break; } point next; for(int i=0;i<=9;i++) { next.x=now.x/10; next.x*=10; next.x+=i; next.y=now.y+1; if(next.x<1000||next.x>=10000) continue; if(vis[next.x]==1) continue; if(next.x==m) { flag1=next.y; break; } if(primes[next.x]==1) { //cout<<next.x<<endl; vis[next.x]=1; q.push(next); } } for(int i=0;i<=9;i++) { int temp=now.x%10; next.x=now.x/100; next.x*=100; next.x+=i*10; next.x+=temp; if(next.x<1000||next.x>=10000) continue; if(vis[next.x]==1) continue; if(next.x==m) { flag1=next.y; break; } if(primes[next.x]==1) { //cout<<next.x<<endl; vis[next.x]=1; q.push((point){next.x,now.y+1}); } } for(int i=0;i<=9;i++) { int temp=now.x%100; next.x=now.x/1000; next.x*=1000; next.x+=i*100; next.x+=temp; if(next.x<1000||next.x>=10000) continue; if(vis[next.x]==1) continue; if(next.x==m) { flag1=next.y; break; } if(primes[next.x]==1) { //cout<<next.x<<endl; vis[next.x]=1; q.push((point){next.x,now.y+1}); } } for(int i=0;i<=9;i++) { int temp=now.x%1000; next.x=now.x/10000; next.x*=10000; next.x+=i*1000; next.x+=temp; if(next.x<1000||next.x>=10000) continue; if(vis[next.x]==1) continue; if(next.x==m) { flag1=next.y; break; } if(primes[next.x]==1) { // cout<<next.x<<endl; vis[next.x]=1; q.push((point){next.x,now.y+1}); } } if(flag1>0) break; q.pop(); } printf("%d\n",flag1); } return 0; }
CD0J/POJ 851/3126 方老师与素数/Prime Path BFS
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原文地址：http://www.cnblogs.com/qscqesze/p/4223350.html
